The general displacement of a simple harmonic oscillator is x =...

Given here, displacement $x = Asin\left(\omegat\right)$

Now, potential energy of a simple harmonic oscillator is $U = \dfrac{1}{2}m{\omega}^{2}{x}^{2} = \dfrac{1}{2}m{\omega}^{2}{A}^{2}\sin^{2}\left(\omegat\right)$,

So, slope of potential energy and time curve is $\dfrac{dU}{dt} = \dfrac{1}{2}\dfrac{m{\omega}^{3}{A}^{2}}{2}\sin 2\omegat$

Now, {\left(\dfrac{dU}{dt}\right)}_\max will be maximum when $\sin 2\omegat = 1$

$⇒ 2\omegat = \dfrac{\pi}{2} ⇒ 2\left(\dfrac{2\pi}{T}\right)t = \dfrac{\pi}{2} ⇒ t = \dfrac{T}{8} ⇒ \beta = 8$